A rocket with total mass 3.03 × 105 kg is in circular orbit around the Earth. It

Question

A rocket with total mass 3.03 × 105 kg is in circular orbit around the Earth. It begins to accelerate at 35.8 m/s2 tangent to its orbit (hence doing no work against gravity). (a) If the speed of the exhausted geses is 4.40 k 10 m/s, at what Fate is the rocket initially burning fuel? kg/s (b) If the rocket were to be launched vertically from Earthi's surface with the same initial acceleration, at what rate would the fuel have to be burned? (Disregard the reduction in exhaust Speed due to the ambient atmospheric pressure.) kg/s

Explanation / Answer

Given Data

Mass of rocket is (m) = 3.03 x 105 Kg

Acceration of rocket (dv/dt), ( a ) = 35.8 m/s2

Exhausted gas velocity is (v) = 4.40 x 103 m/s

Solution:-

a)

But, Accelation in rocket (dv/dt) = (v/m)(dM/dt)

Now rate of fuel burning by ingoring gravitional force is (dM/dt) = -(m/v)*(dv/dt)

                                                                               = (3.03 x 105/4.40 x 103)x a

                                                                               = (303x35.8)/4.40

                                                                               = 2465.32 Kg/s

(b) Fnet = Ft - mg

m*a = Ft - mg

Ft = m*(g+a)

v*(dM/dt) = m*(g+a)

(dM/dt) = m/v*(g+a)

           = (3.03 x 105/4.40 x 103) * (9.8 + 35.8)

           = 3140.2 Kg/s

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